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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Solution:</dfn> <span class="process-math">\(x=2\)</span> is an ordinary point. According to the theorem, we have the following general solution:</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_5.html">
\begin{equation*}
y=\sum_{n=0}^{\infty} a_n (x-2)^n=a_0 y_1(x)+a_1 y_2(x).
\end{equation*}
</div>
<p class="continuation">Substituting it into the ODE,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_5.html">
\begin{equation*}
\begin{aligned}
&amp; 2 \sum_{n=0}^{\infty} n (n-1) a_n (x-2)^{n-2}+(x-2+3) \sum_{n=0}^{\infty}n  a_n (x-2)^{n-1}+3 \sum_{n=0}^{\infty} a_n (x-2)^n=0,\\
&amp;\to 2 \sum_{k=-2}^{\infty} (k+2) (k+1) a_{k+2} (x-2)^k+\sum_{n=0}^{\infty} n a_n (x-2)^n+3 \sum_{n=0}^{\infty} n a_{n} (x-2)^{n-1}+3 \sum_{n=0}^{\infty} a_n (x-2)^n=0,\\
&amp;\to 2 \sum_{n=0} (n+2) (n+1) a_{n+2} (x-2)^n++\sum_{n=0}^{\infty} n a_n (x-2)^n+3 \sum_{n=0}^{\infty} (n+1) a_{n+1} (x-2)^n+3 \sum_{n=0}^{\infty} a_n (x-2)^n=0,\\
&amp;\to \sum_{n=0}^{\infty} \left[    2 (n+2)(n+1) a_{n+2} +3(n+1) a_{n+1}+(n+3) a_n
\right] (x-2)^n=0,
\end{aligned}
\end{equation*}
</div>
<p class="continuation">which finally gives</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_5.html">
\begin{equation}
2(n+2)(n+1) a_{n+2}+3(n+1) a_{n+1}+(n+3) a_n=0.\tag{5.3.2}
\end{equation}
</div>
<p class="continuation">According to the theorem,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_5.html">
\begin{equation*}
y_1(x)=\sum_{n=0}^{\infty} a_n (x-2)^n,
\end{equation*}
</div>
<p class="continuation">with <span class="process-math">\(a_1=0, a_0=1\text{.}\)</span> In (<a href="" class="xref" data-knowl="./knowl/eq5_5.html" title="Equation 5.3.2">(5.3.2)</a>), let <span class="process-math">\(a_1=0, a_0=1\text{.}\)</span> Then</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_5.html">
\begin{equation*}
\begin{aligned}
&amp; n=0: 2 \cdot 2 \cdot 1\cdot  a_2+3 \cdot 1 \cdot 0 +3 \cdot 1=0 \to a_2=-\frac{3}{4},\\
&amp;n=1: 2 \cdot 3 \cdot 2 \cdot a_3+3 \cdot 2 \cdot a_2+4 \cdot 0=0 \to a_3=-\frac{1}{2} a_2=\frac{3}{8},\\
&amp;n=2: 2 \cdot 4 \cdot 3 \cdot a_4+3 \cdot 3 \cdot \frac{3}{8}+6\cdot \left(-\frac{3}{4}\right)=0 \to a_4=\frac{3}{64}.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Thus,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_5.html">
\begin{equation*}
y_1(x)=1-\frac{3}{4}(x-2)^2+\frac{3}{8}(x-2)^3+\frac{3}{64}(x-2)^4+\cdots
\end{equation*}
</div>
<span class="incontext"><a href="sec5_3.html#p-218" class="internal">in-context</a></span>
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